Question

determine a razao de uma P.A. cujo primeiro termo a1=7,65 ,o numero de termos e igual a 18 e seu ultimo termo e igual a 47,77 .
na P.A ( 11,a,5,b,-1,c,-7) , determine :
a)os valores de a,b e c.
b)a razao dessa P.A
c)a soma dos primeiros termos dessa P.A

Answer 1

Resposta:

$1)\ r=2,36\\\\2)\ a)\ a=8,\ b=2,\ c=-4\\\\b)\ r=-3\\\\c)\ S_{7} =14$

Explicação passo a passo:

$a_{1} =7,65\\\\n=18\\\\a_{n} =a_{18} =47,77\\\\r=?$

$a_{1} +(n-1)\times r=a_{n}\\\\(n-1)\times r=a_{n}-a_{1}\\\\r=\frac{a_{n}-a_{1}}{n-1}$

$r=\frac{47,77-7,65}{18-1}\\\\r=\frac{40,12}{17}\\\\r=2,36$

$----------$

$(11,a,5,b,-1,c,-7)\\\\a_{1} =11\\a_{3} =5\\\\a_{1} +(n-1)\times r=a_{n}\\\\11 +(3-1)\times r=a_{3}\\\\11 +2r=5\\\\2r=5-11\\\\2r=-6\\\\r=\frac{-6}{2} \\\\r=-3$

$a_{n} =a_{n-1} +r\\\\a_{2} =a_{2-1} +(-3)\\\\a =a_{1} -3\\\\a=11-3\\\\a=8\\\\\\a_{4} =a_{4-1} +r\\\\b =a_{3} +(-3)\\\\b =5 -3\\\\b=2\\\\a_{6} =a_{6-1} +r\\\\c =a_{5} +(-3)\\\\c=-1-3\\\\c=-4$

$S_{n} =\frac{(a_{1}+a_{n} )\times n}{2} \\\\S_{7} =\frac{(11+(-7) )\times 7}{2}\\\\S_{7} =\frac{(11-7 )\times 7}{2}\\\\S_{7} =\frac{4\times 7}{2}\\\\S_{7} =\frac{28}{2}\\\\S_{7} =14$